Question

Q 1: If the fluorescence signal from a PMT is 0.531 volts, what will the voltage output be if the following occurs:

a. the lamp intensity doubles
b. the concentration is reduced by 1/2
c. pathlength is doubled
d. the quantum yield drops from 0.50 to 0.10

e.temperature increases

Q 2: Predict how the fluorescence signal changes if the following occurs. Explain your answer.

[1] the sample temperature decreases

[2] the excitation light intensity increases by 25%

Answer 1

Solution.

The fluorescence intensity, F is proportional to the amount of light absorbed and the fluorescence quantum yield, Φ

F=kIΦ[εbc]

where k is a proportionality constant, I is the incident light intensity, is the molar absorptivity, b is the path length, and c is the concentration of the substrate.

We are interested in voltage, it will change the proportionality constant only.

a) If the lamp intensity doubles, I₂/I₁ = 2, and F₂/F₁ = 2. F₁ = 0.531; F₂ = 2*0.531 = 1.062 V.

b) If the concentration is reduced by 1/2, c₂/c₁ = 1/2; F₂/F₁ = 1/2. F₁ = 0.531; F2 = 0.531/2 = 0.266 V.

c) If the pathlength is doubled, b₂/b₁ = 2; F₂/F₁ = 2. F₁ = 0.531; F₂ = 2*0.531 = 1.062 V.

d) If the quantum yield drops from 0.50 to 0.10, Φ₂/Φ₁ = 0.1/0.5 = 0.2; F₂/F₁ = 0.2. F₁ = 0.531; F2 = 0.2*0.531 =0.106 V

e) The fluorescence intensity decreases with increasing temperature, so voltage will be less than 0.531 V.

[1] If the sample temperature decreases, the  fluorescence intensity increases so voltage will be greater than 0.531 V.

[2] If the excitation light intensity increases by 25%,  I₂/I₁ = 1.25, and F₂/F₁ = 1.25. The fluorescence intensity increases by 25% as well.